This issue came up in a number theory lecture today. Let $K$ be a number field and let $L/K$ be an abelian (finite Galois) extension. Then there exists a primitive $m$th root of unity $\zeta_m$ so that $K(\zeta_m)\cap L=K$ so that $m$ satisfies a number of nice qualities.

We tried to apply the following fact: if $(m,p)=1$ for every prime $p\in\mathbb Z$ ramifying in $\mathcal O_L$, then $\mathbb Q(\zeta_m)\cap L=\mathbb Q$.

The issue in moving this fact up to $K/\mathbb Q$ is that, for general fields $E_1,E_2,E_3$, $E_1(E_2\cap E_3)\neq E_1E_2\cap E_1E_3$. However in our case, we do have that $K(\zeta_m)\cap L=K$, where $m$ is chosen via primes ramifying from the base field $\mathbb Q$. Perhaps if $K/\mathbb Q$ were Galois, this might be easier, but the question is this:

Given three fields $F,K,L$ contained in some larger field $M$, under what (minimal) conditions do we have $F(K\cap L)=FK\cap FL$?


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